Disclaimer. My resolutions for this set are even "live"r than most. I wrote down my own processes and thoughts as I went, rather than predicting all of the alternatives you might try. Clearly, your resolutions will differ from some of these.
0. Gender-based Group Learning
There are many possible reasons for this. A simple one is that "women are smarter and therefore raise the curve in any group" (yes, there are ways to argue the reverse, but it's more difficulty). Another is that the different groups have significantly different characters and the way men and women behave in different types of groups affects their learning. (You can fill in their behaviors.)
How would we tell? If we make a claim about behavior, we can observe the groups and see if our observations match the claim. Clearly, we would want to observe a wide variety of groups in a wide variety of situations.
1. A Lifetime in Math
I started my consideration of this problem by drawing a timeline.
| childhood | youth | unmarried | childless | parent | +----------------------------------------------------+ |son dies
I then reread the problem and realized that some of the numbers didn't make sense, particularly as I listened to my quite able students discussing the problem. I checked my typing and realized that I had mistyped the amount of time in childhood. Sorry ...
Okay, on to defining variables.
Yuck, that's a lot of variables. On to the fun equations.
Aha! Doing without equations..
Okay, I'll assume that the person who wrote this problem didn't want to work with fractional ages. Hence, her age has to be a multiple of both seven and twelve. If I had to guess, I'd guess 84, which is 7 times twelve. How can I check this? It might help to have the equations, so I'll continue with those equations.
Check! More equations.
First, I'll resolve the two equations using S so that I can eliminate S. Now, we know that D/2 = P-4 or that P = D/2+4 (which is a good form, since everything else is in terms of D).
Next, I'll solve that big sum for D (which shouldn't be so bad, since the other equations are in terms of D).
Isn't that ugly? Well, I guess I'll find a common denominator. Hey! It's 84.
Okay, now I can simplify a little.
Not much of a surprise, but a useful answer.
Reflect. That was a lot of work for a not very intersting (and potentially obvious) answer. As in some previous word problems, I found it more interesting to reflect on the "hidden" values in the problem and solve with them, rather than to work out the formulae. However, I did not a problem in the problem when I tried to work out the formulae.
2. Three More Generations
Oh boy, another one of these aging problems. Well, let's see.
Let's see, we'll express I in terms of M in the first equation and D in terms of M in the last equation and then plug into the middle equation.
Hmmm ... Not a very useful result. Stuck!
Let's think about those initial statements again. Aha! Wait a minute, if my mother is twice as old as I am and eight times as old as my daughter, then I must be four times as old as my daughter. That means that I already knew that the ratio of her age to my age was half the ration of my age to my daughter's age! I only have two useful equations and three unknowns, so the problem isn't solvable for particular values.
In fact, we can choose a large number of ages that meet these criteria.
Check. Yes, all of those examples meet the criteria.
Reflect. Although this was similar to a number of "how old are we" problems, the author of this problem (SamR) either didn't try to ensure that the equations were independent or purposefully wrote the question so that they didn't give sufficient information (okay, I wrote them, I know that it's the latter). Did I learn something from this: yes, sometimes the same technique does not work the same way for all similar problems. I also learned that the phrasing of problems is important.
3. Selling Candy
Oh boy, more numeric problems. 9,000 have nuts and/or chocolate. We'll divide the world once more.
On to equations:
Plug in 7200 for C+Y in the third equation and we get
Plug in 1800 for C in the first equation and we get
Plug in 4000 for Y in the second equation and we get
Reflect This was very similar to the sandwiches problem from problem set 1.
4. The Fake Coin
We did this as a group for nine coins.
5. Campus Bikes
In order to understand the long term effect, we should choose a number of years that will give us even numbers. Since we have seen issues of ten years and fifteen years, we might as well look at thirty years. Does it matter how many bikes we buy? Apparently not, since we'll just multiply by the number of bikes. Does it matter how much money we start out with? If we start out with too little, it's a problem. However, the endowment is big enough that we don't need to worry about it. If we have some extra money in both cases, that's okay, too. We'll start with $10,000, since that's obviously enough to purchase a few generations of bikes.
Case one: no Zoomcoat, no interest, thirty years. In year 0, we buy one Zoomster, leaving us with $8,900. In year 10, we sell that Zoomster for $100, giving us $9,000. We then buy another bike, leaving us with $7,900. In year 20, we sell that Zoomster for $100, giving us $8,000. We then buy another one for $1,100, leaving us with $6,900. In year 30, we sell that for $100, leaving us with $7,000.
Case two: Zoomcoat, no interest, thirty years. In year 0, we buy one Zoomster and Zoomcoat, leaving us with $8,600. In year fifteen, we buy another Zoomster and Zoomcoat, leaving us with $7,200. This looks like a win.
Observe. I note that $3,000 would have been enough to start with. Will that make analysis of the problem easier? Perhaps, but not enough so to warrant many changes.
Reflect. We have an answer in a no-interest world, but it seems as if we might make more money by investing that $300, especially since some of it will be invested for a long time. How much do I make? In ten years, I make principal*(1.05)^10 or approximately principal*1.63. In fifteen years, I make principal*(1.05)^15 or approximately principal*2.08.
Case three: no Zoomcoat, interest, thirty years. In year 0, we buy one Zoomster, leaving us with $8,900. In year 10, we have collected approximately $5,607 in compound interest, giving us $14,507. We also sell the Zoomster for $100, giving us $14,607. We buy another Zoomster for $1,100, leaving us with $13,507. In year 20, we have collected approximately $8,509 in compound interest, giving us $22,016. We sell the Zoomster for $100, giving us $22,116. We then buy another for $1,110, leaving us with $21,006. Finally, in year 30, we have collected approximately $13,233 in interest, giving us $34,239. We sell the remaining Zoomster for $100, and end up with $34,339. Not a bad chunk of change.
Case four: Zoomcoat, interest, thirty years. In year 0, we buy one Zoomster with Zoomcoat for $1,400, leaving us with $8,600. In year 15, we have collected approximately $9,288 in compound interest, giving us $17,888. We then buy another Zoomster with Zoomcoat, leaving us with $16,488. In year 30, we have collected approximately $17,807 in compound interest, giving us $34,295.
Check. Um, I did lots of rounding in this problem, and it's likely that the bank would, too. I'm not sure whether or not my answer is correct. If pressed, I'd choose to do without Zoomcoat, but wouldn't argue strenuously in either direction. I might also want to try the same problem with different starting amounts. It seems with a higher interest rate I might do a little bit better.
Reflect. Not a very satisfactory solution, although I found the problem interesting as it was a little bit more "real world" than some of the other problems. I did notice that even at 5%, you can make a lot of money on your money after thirty years.
Aha!. There does seem to be another way to think about the problem that helps us ignore some of the problematic issues. If I don't buy Zoomcoat then the amount extra I have to invest depends on which year it is. That is, rather than considering amount remaining in my account, I'll consider the extra amount I have for each strategy. I might be able to use this to compute something useful.
Now, how does this extra cash affect our income?
In the "no Zoomcoat" case.
In the "Zoomcoat" case.
Does this help? At least I can come up with numbers that better describe the differential.
It does seem that skipping the Zoomcoat is better.
Check. If my reasoning is sound, the differences between the numbers in both cases should be about the same. $2,851 - $2,435 = $416. $34,339 - $34,295 = $44. Um. That's not good.
6. Differential Taxes
This seems like a good problem in which to write formulae. Let I be my untaxed (gross) income. My real (net) income will be I * (1-(I/5000)/100). or (-1/500,000)*I^2 + I. That's a quadratic equation describing a parabola that "points downwards". I know that parabolas are symmetrical, so the high point on this parabola will be midway between the two zeroes of the function.
In A*I^2 + B*I + C form,
I could then find the zeroes of the function by using the legendary formula (b*b +- 4ac)/2a. However, since this is such a simple function, I can also write it as I((-1/500,000)I + 1). This is 0 when I is 0 and when I is 500,000. Hence, I take home the most money when my gross income is midway between those two values. I should strive to earn $250,000.
Check. I'll try a few examples to see if my answer seems reasonable. If I make $250,000, I take home 50% of $250,000, or $125,000. If I make $200,000, I take home 60% or $120,000. If I make $300,000, I take home 40% or $120,000.
Reflect. What does this do to Bill Gates? More realistically, it suggests that this taxation system caps people's real income, at a number that will feel high to many Americans and low to a few Americans. It might be fun to look up incomes in the census and see what it really does (and whether it would work).
7. The Bricklayers
Okay, time for more variables and formulae. We'll start with variables.
The first bricklayer takes 9 hours to build the wall.
The second bricklayer takes 10 hours to build the wall.
The two bricklayers can build the wall in 5 hours, but they two bricklayers lay 10 fewer bricks per hour.
Three equations, three unknowns. Looks good. Let's rephrase the first two equations so that we have A in terms of W and B in terms of W.
We can now work with the third.
Stuck! That's a pretty meaningless result. How can a wall have negative bricks?
Check. Okay, it appears that I can't do math. 5*19 = 95, rather than 85. Let's try again.
Reflect. I'm clearly too confident in my "simple math" skills and pick answers without reflecting sufficiently much. This is yet another "determine variables and formulae and solve" problem. It was a little bit more interesting in that the variables I picked had to do with rates, but that didn't seem to be a big difference.
8. The City Gates
This is one of those "what's really being asked" questions. Clearly, the person who wrote it was trying to rephrase the "St. Ives" question. In one sense, this version doesn't give us enough data. How many were at the city gates? Well, we only know about the people who were described in the question.
This seems to be another one in which it helps to have variables.
The total number of "things" at the city gates is 1 + 1 + M + S + C + K. The first one is "me". The next one is "someone".
That totals 2802.
Check. Should I be counting sacks? It does say "kits, cats, sacks, and mates. Is there any potential tricks in the problem. Well, it is possible that the mates shared the sacks (it doesn't say "each mate had seven different sacks"), so it could be possible that there are only seven sacks. For example, if I heard "I just met Sam and Michelle. Each of them has a son. How many son's did they have?" It's possible that they have two. It's also possible that they have one. Clearly, this problem is phrased somewhat confusingly.
Reflect. I've always found "St. Ives" a little bit silly. This one has less of a trick, but could also have hidden tricks. "It depends" would be a reasonable answer. "Who cares" would also be a reasonable answer.
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